题意:给一个序列以及\(n\)个查询,每一个查询是问(假装)把第\(a_i\)个数改为\(b_i\)之后原序列的最长上升子序列的长度. 思路:线段树优化\(dp\). 肯定离线做啊. 首先我们考虑\(dp\)的状态是\(dp_L(i)\)表示以第\(i\)位为结束的最长上升子序列的长度和方案数\(mod\ 998244353\). 之所以方案数要模某个数是因为这个方案数太大太大了,肯定爆\(long\ long\). 就因为这个我\(wa\)了一次\(test\ case\ 13\). 开心的不…

转载请注明出处:http://www.cnblogs.com/Delostik/p/3553114.html 目前已有[A B C D E] 例行吐槽:趴桌子上睡着了 [A. Genetic Engineering] http://codeforces.com/contest/391/problem/A 问最少插入几个字符,使得字符串不存在连续偶数个相同字母.不说什么 #include <iostream> #include <string> using namespace std…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of b consecu…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Vasya is currently at a car rental service, and he wants to reach cinema. The film he has bought a ticket for starts in t minutes. There is a str…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Famous Brazil city Rio de Janeiro holds a tennis tournament and Ostap Bender doesn't want to miss this event. There will be n players participat…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and th…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization…

[链接] 我是链接,点我呀:) [题意] [题解] 分类讨论一波 设第一个数组的奇数个数为cnt1 第二个数组的奇数个数为cnt2 显然只有在(cnt1+cnt2)%2==0的情况下. 才可能第一个数组的和为偶数,第二个数组的和也为偶数 (因为奇数都要出现偶数次才可以. 所以只可能cnt1和cnt2都是偶数,那么输出0 否则,cnt1和cnt2都是奇数,看看有没有一个位置i只有a[i]或只有b[i]是奇数.有的话输出1. 其他情况都非法. [代码] #include <bits/stdc++.h…

                                                        C. An impassioned circulation of affection                                                                time limit per test  2 seconds                                                          …